【LeetCode】Two Sum II - Input array is sorted

【Description】

Given an array of integers that is already sorted in ascending order, find two numbers such that they add up to a specific target number.

The function twoSum should return indices of the two numbers such that they add up to the target, where index1 must be less than index2.

Note:

• Your returned answers (both index1 and index2) are not zero-based.
• You may assume that each input would have exactly one solution and you may not use the same element twice.

Example:

Input: numbers = [2,7,11,15], target = 9
Output: [1,2]
Explanation: The sum of 2 and 7 is 9. Therefore index1 = 1, index2 = 2.

【AC code】

一、暴力法　　时间复杂度：O(n^2)

 class Solution {
     public int[] twoSum(int[] numbers, int target) {
         int arrlen = numbers.length;
         for (int i = 0; i < arrlen - 1; i++) {
             for (int j = i + 1; j < arrlen; j++) {
                 if (numbers[i] + numbers[j] == target) return new int[]{i + 1, j + 1};
             }
         }
         return new int[]{};
     }
 }

二、二分查找法　　时间复杂度：O(nlogn)

 class Solution {
     public int[] twoSum(int[] numbers, int target) {
         int arrlen = numbers.length;
         for (int i = 0; i < arrlen; i++) {
             int left = i + 1, right = arrlen - 1;
             while (left <= right) {
                 int mid = left + (right - left) / 2;
                 int tmp = numbers[i] + numbers[mid];
                 if (tmp > target) right = mid - 1;
                 else if (tmp < target) left = mid + 1;
                 else return new int[]{i + 1, mid + 1};
             }
         }
         return new int[]{};
     }
 }

三、双索引法　　时间复杂度：O(n)

 class Solution {
     public int[] twoSum(int[] numbers, int target) {
         int left = 0, right = numbers.length - 1;
         while (left < right) {
             int tmp = numbers[left] + numbers[right];
             if (tmp == target) return new int[]{left + 1, right + 1};
             else if (tmp > target) right--;
             else left++;
         }
         return new int[]{};
     }
 }